Let $f: U \longrightarrow \mathbb{R}^{m}$ differentiable in the open set $U \subset \mathbb{R}^{m}$. If, for a $b \in \mathbb{R}^{n}$, the set $f^{-1}(b)$ has an accumulation point $a \in U$. Then $f'(a): \mathbb{R}^{m} \longrightarrow \mathbb{R}^{n}$ is not injective.
I have no ideia to start this question. I know that I have a sequences $(x_{k})$ with $x_{i} \neq x_{j}$ if $i \neq j$ such that $(x_{k}) \rightarrow a$. But I don't know how to use it. I don't want a solution, just a hint.
On
Assume that $(a_{n})\subseteq f^{-1}(b)$ is such that $a_{n}\rightarrow a$, $a_{n}\ne a$, then $f(a_{n})=b$. Since $f$ is continuous at $x=a$, we have $f(a_{n})\rightarrow f(a)$, so $b=f(a)$.
Given $\epsilon>0$, we have for some $\delta>0$ that $|f(a+h)-f(a)-f'(a)h|<\epsilon|h|$ for $0<|h|<\delta$. Realizing $h=a_{n}-a$, then for large $n$, we have $|f(a_{n})-f(a)-f'(a)(a_{n}-a)|<\epsilon|a_{n}-a|$, so $|b-f(a)-f'(a)(a_{n}-a)|<\epsilon|a_{n}-a|$ and so $|f'(a)(u_{n})|<\epsilon$ for $u_{n}=\dfrac{a_{n}-a}{|a_{n}-a|}$. Note that $u_{n}\in\{|x|=1\}$ where $\{|x|=1\}$ is sequentially compact, so have some convergent subsequence, also denoting by $(u_{n})$ that $u_{n}\rightarrow u\in\{|x|=1\}$.
By taking $n\rightarrow\infty$ to $|f'(a)(u_{n})|<\epsilon$, we have $|f'(a)(u)|\leq\epsilon$. This is true for all $\epsilon>0$, so $f'(a)(u)=0$.
You can manage to arrange a sequence $h_n$ such that $h_n \to 0$ and $$f(a+h_n)=f(a). $$ Since $$f(a+h_n)=f(a)+Df_a \cdot h_n +\epsilon(h_n),$$ you know that $$Df_a \cdot h_n+\epsilon(h_n)=0.$$ For all $n$. Divide by $\Vert h_n\Vert$ and use compactness of the sphere to get an element $v \in S^{n-1}$ such that $Df_a \cdot v =0$.